Prime ideals and irreducible ideals

Assume that $J\neq R$ is prime, and $J=I_1\cap I_2$. Then $I_1I_2\subseteq I_1\cap I_2 = J$; since $J$ is prime, then $I_1\subseteq J$ or $I_2\subseteq J$. But then, if we have $I_1\subseteq J = I_1\cap I_2$, so $I_1\subseteq I_2$; in particular, $J=I_1\cap I_2 = I_1$, so we conclude that $J=I_1$; analogously, if $I_2\subseteq J$, then we conclude that $J=I_2$, so the condition for irreducibility is satisfied. You don't need $R$ to be an integral domain. (I didn't even need $R$ to be commutative, since the proof above is ideal-wise, not element-wise!)

Take $R=\mathbb{Z}$, $J=(p^n)$ where $p$ is a prime and $n\gt 0$. If $(p^n)=(a)\cap (b) = (\mathrm{lcm}(a,b))$ then we must have $a=\pm p^i$, $b=\pm p^j$ for some $i,j$, $0\leq i,j\leq n$; and we must have $\max\{i,j\}=n$. Hence $(p^n)=(a)$ or $(p^n)=(b)$, so $(p^n)$ is irreducible. However, it is only prime if $n=1$.

Is incorrect, as witnessed by the example in 2; even Euclidean is not sufficient for irreducible to imply prime.

But in a PID (or more generally, a Noetherian ring), irreducible and radical does imply prime. For in a PID (or as noted by user23214, in a Noetherian ring), irreducible implies primary; if $J$ is primary and radical, then it is prime. But this is probably too much of a sledgehammer condition.

(An ideal $I$ is primary if $I\neq R$ and $xy\in I$ implies $x\in I$ or $y^n\in I$ for some $n\gt 0$; an ideal $I$ is radical if $I=\sqrt{I}$, if $x^n\in I$ for some $n\gt 0$ implies $x\in I$.)

Same problem for the statement about irreducible elements/ideals. But in a PID, an ideal generated by an irreducible is necessarily irreducible (if $(m)=(a)\cap(b)$, then $a|m$, and irreducibility of $m$ implies $(a)=(1)$ or $(a)=(m)$; similarly with $b$, and we cannot have both $(a)=(1)$ and $(b)=(1)$), though the converse need not hold, as witnessed by the example in 2.